A 2.5 F air capacitor is connected to a 6 V battery and charged. (a) What is the charged acquired by the capacitor? (b)How much energy is stored in the capacitor?
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- Find (a) the equivalent capacitance of the capacitors in Figure, (b) the charge on each capacitor, and (c) the potential difference across each capacitor. 6.00HF 8.00 pF = 2.00 uF E8.00 uF 9.00 V7. (a) The space between two parallel plate capacitor is filled with two dielectric materials of same dimensions. What is the resulting capacitance as a function of K₁, K₂ and the original Co when the space in between is empty? (b) What would be the resulting capacitance of the following setup then? Express your answer as a function of K₁, K₂ and the original Co when the space in between is empty?Two capacitors, C, = 6.00 µF and C, = 11.0 µF, are connected in parallel, %3D and the resulting combination is connected to a 9.00-V battery. (a) Find the equivalent capacitance of the combination. µF (b) Find the potential difference across each capacitor. V1 = V2 = v (c) Find the charge stored on each capacitor. HC
- A 15-μF capacitor is charged to 40 V and then connected across an initially uncharged 25-μF capacitor. What is the final potential difference across the 25-μF capacitor?The capacitance of an empty capacitor is 1.2 μF. The capacitor is connected to a 12-V battery and charged up. With the capacitor connected to the battery, a slab of dielectric material is inserted between the plates. As a result, 2.6 × 10–5 C of additional charge flows from one plate, through the battery, and onto the other plate. What is the dielectric constant of the material?